What is the integral of (x+1)/(x(x^2+2)) ?

The integral of (x+1)/(x(x^2+2)) is 1/(sqrt(2))arctan(x/(sqrt(2)))-1/4 (ln|x^2+2|-2ln(x))+C

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integral of (x+1)/(x(x^2+2))

\int \frac{x + 1}{x ( x ^{2} + 2 )} d x

\frac{1}{2} arctan (\frac{x}{2}) - \frac{1}{4} (ln x^{2} + 2 - 2 ln (x)) + C

Solution steps

\int \frac{x + 1}{x ( x ^{2} + 2 )} d x

Expand

\frac{x + 1}{x ( x ^{2} + 2 )} : \frac{x}{x ( x ^{2} + 2 )} + \frac{1}{x ( x ^{2} + 2 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{x ( x ^{2} + 2 )} d x + \int \frac{1}{x ( x ^{2} + 2 )} d x

\int \frac{x}{x ( x ^{2} + 2 )} d x = \frac{1}{2} arctan (\frac{x}{2})

\int \frac{1}{x ( x ^{2} + 2 )} d x = - \frac{1}{4} (ln x^{2} + 2 - 2 ln (x))

= \frac{1}{2} arctan (\frac{x}{2}) - \frac{1}{4} (ln x^{2} + 2 - 2 ln (x))

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x}{2}) - \frac{1}{4} (ln x^{2} + 2 - 2 ln (x)) + C

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What is the integral of (x+1)/(x(x^2+2)) ?
The integral of (x+1)/(x(x^2+2)) is 1/(sqrt(2))arctan(x/(sqrt(2)))-1/4 (ln|x^2+2|-2ln(x))+C